Asymptotics for $g(n) = \sum_{k = 1}^{n - 1} {\frac{{\log (1 + p_k)}}{p_k}}$?

Question

Let $p_k$ be the $k$th prime.

Now define $g(n)$ as

$$g(n) = \sum_{k = 1}^{n - 1} {\frac{{\log (1 + p_k)}}{p_k}}$$

What are the asymptotics for this $g(n)$ ?

The related sum

$$ \sum_{k = 1}^{n - 1} {\frac{{\log (1 + k)}}{k}}$$ has been well approximated by user gary ( https://math.stackexchange.com/users/83800/gary) in the analogue question

Asymptotics for $ f(n) = \prod_{k=2}^{n} \sqrt[k-1]{k}$?

I copy his answer :

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By the Abel–Plana formula \begin{align*} \sum\limits_{k = 1}^{n - 1} {\frac{{\log (1 + k)}}{k}} = \int_1^{n - 1} {\frac{{\log (1 + t)}}{t}{\rm d}t} & - 2\int_0^{ + \infty } {\frac{1}{{{\rm e}^{2\pi y} - 1}}{\mathop{\rm Im}\nolimits}\! \left( {\frac{{\log (2 + {\rm i}y)}}{{1 + {\rm i}y}}} \right){\rm d}y} \\ & + \frac{{\log 2}}{2} + \mathcal{O}\!\left( {\frac{{\log n}}{n}} \right). \end{align*} Here $$ - 2\int_0^{ + \infty } {\frac{1}{{{\rm e}^{2\pi y} - 1}}{\mathop{\rm Im}\nolimits} \!\left( {\frac{{\log (2 + {\rm i}y)}}{{1 + {\rm i}y}}} \right){\rm d}y} = \int_0^{ + \infty } {\frac{1}{{{\rm e}^{2\pi y} - 1}}\frac{{y\log (4 + y^2 ) - 2\arctan (y/2)}}{{y^2 + 1}}{\rm d}y} $$ and $$ \int_1^{n - 1} {\frac{{\log (1 + t)}}{t}{\rm d}t} = \frac{\log ^2 n}{2} + \frac{{\pi ^2 }}{{12}} + \mathcal{O}\!\left( {\frac{{\log n}}{n}} \right). $$ Accordingly, $$ \prod\limits_{k = 2}^n {\sqrt[{k - 1}]{k}} = Cn^{\frac{1}{2}\log n} \left( {1 + \mathcal{O}\!\left( {\frac{{\log n}}{n}} \right)} \right) $$ as $n\to +\infty$, where $$ \log C = \frac{{\pi ^2 }}{{12}} + \frac{{\log 2}}{2} + \int_0^{ + \infty } {\frac{1}{{{\rm e}^{2\pi y} - 1}}\frac{{y\log (4 + y^2 ) - 2\arctan (y/2)}}{{y^2 + 1}}{\rm d}y} = 1.18493104146 \ldots $$ Addendum. Another way is to note that $$ \sum\limits_{k = 1}^{n - 1} {\frac{{\log (1 + k)}}{k}} = \sum\limits_{k = 1}^{n - 1} {\frac{{\log k}}{k}} + \sum\limits_{k = 1}^{n - 1} {\frac{{\log (1 + 1/k)}}{k}} = \frac{{\log ^2 n}}{2} + \gamma _1 + \sum\limits_{k = 1}^\infty {\frac{{\log (1 + 1/k)}}{k}} + o(1) $$ where $\gamma_1$ is one of the Stieltjes constants. For the infinite series see this answer of mine or $\text{A}131688$ in the OEIS.

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That is a nice application of Abel-Plana. Maybe the other formula from Abel is also useful here :

https://en.wikipedia.org/wiki/Abel%27s_summation_formula

Since we are working over the primes.

The goal or main idea was to use this for number theory, ideas such as the generalizations of Collatz or PNT. See the " heuristic estimate " made here :

Collatz variant $7 x + 1$?

And (more in the context of the PNT related ideas ) the idea of some kind of average lenght of prime factorization based on how often we can devide by a prime.

Those heuristic ideas do assume alot of independance and are not formal yet. But that is where the idea is coming from.

A similar function maybe easier to work with is

$$h(n) = \sum_{p_k < n} {\frac{{\log (1 + p_k)}}{p_k}}$$

Where we just consider the sum over the primes smaller than $n$.

Answer 1

The definition of the von Mangoldt function $\Lambda $ is $$\Lambda (n)=\left \{ \begin {array}{ll}\log p&\text { if }n=\text {power of prime $p$}\\ 0&\text { otherwise.}\end {array}\right .$$ Directly from this definition you get, if $n=\prod _{p|n}p^N$, $$\log n=N\sum _{p|n}\log p=\sum _{p^m|n}\log p=\sum _{d|n}\Lambda (d)=\sum _{ab=n}\Lambda (a)$$ so $$\sum _{n\leq X}\log n=\sum _{ab\leq X}\Lambda (a)=\sum _{a\leq X}\Lambda (a)\underbrace {\sum _{b\leq X/a}1}_{=X/a+\mathcal O(1)}=X\sum _{a\leq X}\frac {\Lambda (a)}{a}+\mathcal O\left (X\right ).$$ On the other hand you can compare the LHS sum here to an integral to get $$\sum _{n\leq X}\log n=X\log X+\mathcal O(X)$$ so we've shown $$\sum _{a\leq X}\frac {\Lambda (a)}{a}=\log X+\mathcal O(1).$$ Directly from the definition of $\Lambda $ this sum is $$\sum _{p^m\leq X}\frac {\log p}{p^m}=\sum _{p\leq X}\frac {\log p}{p}+\underbrace {\mathcal O\left (\sum _{p^m\leq X\atop {m\geq 2}}\frac {\log p}{p^m}\right )}_{\ll \sum _{n=1}^\infty \frac {\log n}{n^2}\ll 1}$$ so we've shown $$\sum _{p\leq X}\frac {\log p}{p}=\log X+\mathcal O(1).\hspace {10mm}(A)$$ This quantity is almost your sum, we just need to sort the $p,p+1$ issue. Since $$\log (p+1)=\log p+\log (1+1/p)=\log p+\mathcal O\left (1/p\right )$$ we get from (A) $$\sum _{p\leq X}\frac {\log (p+1)}{p}=\log X+\mathcal O(1)+\underbrace {\mathcal O\left (\sum _{p\leq X}\frac {1}{p^2}\right )}_{\ll 1}$$ and we're ok.

Answer 2

By Mertens' first theorem, \begin{align*} P(x)& :=\sum\limits_{p \le x} {\frac{{\log (1 + p)}}{p}} = \sum\limits_{p \le x} {\frac{{\log p}}{p}} + \sum\limits_{p \le x} {\frac{{\log (1 + 1/p)}}{p}} \\ & = \left[ {\log x + K + \mathcal{O}({\rm e}^{ - c\sqrt {\log x} } )} \right] + \left[ {\sum\limits_p {\frac{{\log (1 + 1/p)}}{p}} + \mathcal{O}\!\left( {\frac{1}{{x\log x}}} \right)} \right] \\ & = \log x + K' + \mathcal{O}({\rm e}^{ - c\sqrt {\log x} } ) \end{align*} with a suitable $c>0$, and $$ K = - \gamma - \sum\limits_p {\frac{{\log p}}{{p(p - 1)}}} = −1.3325\ldots $$ and $$ K' = - \gamma - \sum\limits_p {\frac{{\log p}}{{p(p - 1)}}} + \sum\limits_p {\frac{{\log (1 + 1/p)}}{p}} = - 0.9485 \ldots, $$ where $\gamma$ is the Euler–Mascheroni constant. Then using the prime counting function $\pi$, $g(n)=P(\pi(n-1))$.